Simple Maths Problem

Here was a simple maths problem that was in a magazine (‘New Scientist’) a few days ago. I post this because I think it is a fairly elegant solution to one of these problems which would normally be attacked by brute force.

The question is: Solve the following equation, where a, b and c are positive integers:

\frac{ab+1}{abc+b+c} = 0.138

The first step is to realise that 0.138 = 69 ÷ 500. Then invert the equation:

\frac{abc+b+c}{ab+1} = \frac{500}{69}

Now we can write this as:

c + \frac{b}{ab+1} = 7 + \frac{17}{69}

As all numbers are positive, b < ab+1. Therefore the fraction that is the second term on the left hand side must be less than 1. Hence c = 7.

Subtract 7 from both sides and multiplying through, we now have that 69b = 17ab + 17. Factorising, we can write that as b(17a-69) = 17

Now 17 is prime, and so one or the other factors must be 1. But it cannot be b, as otherwise 17 | 69, which cannot be true as 69 is 3 × 23. Thus we have 17a – 69 = 1, and so a = 4, and  b = 17. This must be the only solution as all other cases were ruled out.

This can be generalised to other numbers than 0.138, of course, but it doesn’t seem too interesting.

 

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